For the third talk (and a link to the second one), see here. The fourth talk took place on October 12.

We want to show the following version of Theorem 2:

Theorem.Suppose is a singular strong limit cardinal of uncountable cofinality. Then the following are equivalent:

For each ideal on , player II has a winning coding strategy in .

.

Since has uncountable cofinality, option 2 above is equivalent to saying that the instance of corresponding to holds.

Before we begin the proof, we need to single out some elementary consequences in cardinal arithmetic of the assumptions on . First of all, since is singular strong limit, then for any cardinal , we have that

Also, since the cofinality of is uncountable, we have Hausdoff’s result that if , then . I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and here.

We are ready to address the Theorem.

Proof. We use Theorem 1. If option 1. fails, then there is an ideal on with .

Note that , and . Moreover, if , then since, otherwise,

.

So and then, by Hausdorff, in fact , and option 2. fails.

Suppose option 2. fails and let , so and . We use to build an ideal on with .

For this, we use that there is a large almost disjoint family of functions from into . Specifically:

Lemma. If is singular strong limit, there is a family with and such that for all distinct , we have that .

In my notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions , and then replace them with (appropriate codes for) the branches they determine through the tree . Distinct branches eventually diverge, and this translates into the corresponding functions being almost disjoint.

Pick a family as in the lemma, and let be a subfamily of size . Let . We proceed to show that and use to define an ideal on as required.

First, obviously . Since and , it follows that , or else , since is strong limit.

Now define

Clearly, is an ideal. We claim that . First, each singleton with is in , so . Define by . Since the functions in are almost disjoint, it follows that is 1-1. Let be the image of . By construction, is cofinal in . But then

,

where the first inequality follows from noticing that any has size at most . It follows that , as claimed.

Finally, we argue that , which completes the proof. For this, consider a cofinal , and a map such that for all , we have .

Since is cofinal in , it follows that is cofinal in . But this gives the result, because

[This document was typeset using Luca Trevisan‘s LaTeX2WP. I will refer to result (or definition ) from last lecture as ]

A. The Galvin-Hajnal rank and an improvement of Theorem 3.1

Last lecture, I covered the first theorem of the Galvin-Hajnal paper and several corollaries. Recall that the result, Theorem 3.1, states that if and are uncountable regular cardinals, and is -inaccessible, then for any sequence of cardinals such that for all

In particular (see, for example, Corollary 3.7), if and is strong limit, then

The argument relied in the notion of an almost disjoint transversal. Assume that is regular and uncountable, and recall that if is a sequence of sets, then is an a.d.t. for Here, is an a.d.t. for iff and whenever then is bounded.

With as above, Theorem 3.1 was proved by showing that there is an a.d.t. for of size and then proving that, provided that for all then

In fact, the argument showed a bit more. Recall that if then Then, for any ,

The proof of this result was inductive, taking advantage of the well-foundedness of the partial order defined on by iff is bounded in That is well-founded allows us to define a rank for each and we can argue by considering a counterexample of least possible rank to the statement from the previous paragraph.

In fact, more precise results are possible. Galvin and Hajnal observed that replacing the ideal of bounded sets with the nonstationary ideal (or, really, any normal ideal), results in a quantitative improvement of Theorem 3.1. Read the rest of this entry »

In this section I want to present two theorems of Galvin and Hajnal that greatly generalize Silver’s theorem. I focus on a “pointwise” (or everywhere) result, that gives us information beyond the pointwise theorems from last lecture, like Corollary 23. Then I state a result where the hypotheses, as in Silver’s theorem, are required to hold stationarily rather than everywhere. From this result, the full version of Silver’s result can be recovered.

Both results appear in the paper Fred Galvin, András Hajnal, Inequalities for Cardinal Powers, The Annals of Mathematics, Second Series, 101 (3), (May, 1975), 491–498, available from JSTOR, that I will follow closely. For the notion of -inaccessibility, see Definition II.2.20 from last lecture.

Theorem 1.Let be uncountable regular cardinals, and suppose that is -inaccessible. Let be a sequence of cardinals such that for all Then also

The second theorem will be stated next lecture. Theorem 1 is a rather general result; here are some corollaries that illustrate its reach:

Corollary 2.Suppose that are uncountable regular cardinals, and that is -inaccessible. Let be a cardinal, and suppose that for all cardinals Then also

Proof. Apply Theorem 1 with for all

Corollary 3. Suppose that are uncountable regular cardinals, and that is -inaccessible. Let be a cardinal of cofinality and suppose that for all cardinals Then also

Proof. Let be a sequence of cardinals smaller than such that and set for all Then for all by assumption. By Theorem 1, as well.

Corollary 4.Let be cardinals, with and regular and uncountable. Suppose that for all cardinals Then also

Proof. This follows directly from Corollary 2, since is regular and -inaccessible.

Corollary 5. Let be cardinals, with and of uncountable cofinality Suppose that for all cardinals Then also

Proof. This follows directly from Corollary 3 with

Corollary 6.Let be an ordinal of uncountable cofinality, and suppose that for all Then also

Proof. This follows from Corollary 5 with and

Corollary 7. Let be an ordinal of uncountable cofinality, and suppose that for all cardinals and all Then also

Proof. This follows from Corollary 4: If , then by Theorem II.1.10 from lecture II.2. But so both and are strictly smaller than

Corollary 8. If for all then also

Proof. By Corollary 5.

Corollary 9.If for all then also

Proof. By Corollary 7.

Notice that, as general as these results are, they do not provide us with a bound for the size of for the first cardinal of uncountable cofinality that is a fixed point of the aleph sequence, not even under the assumption that is a strong limit cardinal.

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

A simple example is the permutation $\pi$ given by $\pi(n)=n+2$ if $n$ is even, $\pi(1)=0$, and otherwise $\pi(n)=n−2$. It should be clear that $\pi$ is computable and has the desired property. By the way, regarding the footnote: if a bijection is computable, so is its inverse, so $\pi^{-1}$ is computable as well. In general, given a computable bijection $\s […]

The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may a […]

The reflection principle is a theorem schema in ZFC, meaning that for each formula $\phi(\vec x)$ we can prove in ZFC a version of the principle for $\phi$. In particular, it gives us that if $\phi$ holds (in the universe of sets) then there is some ordinal $\alpha$ such that $V_\alpha\models \phi$. It follows from this that (assuming its consistency) $\math […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]